Answer:
c. Observed data points that are far from the other observed data points in the horizontal direction
True, by definition are observed data points that are far from the other observed data points in the horizontal direction.
Step-by-step explanation:
When we conduct a regression we consider influential points by definition "an outlier that greatly affects the slope of the regression line". Based on this case we can analyze one by one the possible options:
a. Observed data points that are close to the other observed data points in the horizontal direction
False. If are close to the other observed values then are not influential points
b. Observed data points that are far from the least squares line
False, that's the definition of outlier.
c. Observed data points that are far from the other observed data points in the horizontal direction
True, by definition are observed data points that are far from the other observed data points in the horizontal direction.
d. Observed data points that are close to the least squares line
False. If are close to the fit regression line adjusted then not affect the general equation for the model and can't be considered as influential points
Let V be the volume of the solid obtained by rotating about the y-axis the region bounded y = 25x and y = x2 25 . Find V by slicing.
Answer:
[tex]volume\ = \pi \frac{625^2}{6}[/tex]
Step-by-step explanation:
See the attached figure.
y₁ = 25x and y₂ =x²
The intersection between y₁ and y₂
25x = x²
x² - 25x = 0
x(x-25) = 0
x = 0 or x =25
y = 0 or y =25² = 625
The points of intersection (0,0) and (25,625)
To find the volume of the solid obtained by rotating about the y-axis the region bounded by y₁ and y₂
y₁ = 25x ⇒ x₁ = y/25 ⇒ x₁² = y²/625
y₂ =x² ⇒ x₂ = √y ⇒ x₂² = y
v = ∫A(y) dy = π ∫ (x₂² - x₁²) dy
∴ V =
[tex]\pi \int\limits^{625}_0 {y-\frac{y^2}{625} } \, dy =\pi( \frac{y^2}{2} -\frac{y^3}{3*625} ) =\pi (\frac{625^2}{2} -\frac{625^3}{3*625}) =\pi ( \frac{625^2}{2}-\frac{625^2}{3}) =\pi \frac{625^2}{6}[/tex]
To find the volume of the solid formed by rotating the region bounded by y = 25x and y = x²25 about the y-axis, one needs to calculate the area of a typical slice, then integrate this area over the range of x-values.
Explanation:To find the volume V of the solid obtained by rotating the region bounded by y = 25x and y = x225 about the y-axis, we can apply the method of slicing.
First, we need to find the area A(x) of a typical slice perpendicular to the x-axis. Here we have two functions, finding the x- values where these functions intersect gives the bounds on the integral for the volume.
We can find the area A(x) = pi*(outer radius)2 - pi*(inner radius)2 = pi[(25x)ˆ2 - (xˆ2 * 25)ˆ2].
To find V, we integrate A(x) over the interval of x-values. This will give us the exact volume of the solid.
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The list of individuals from which a sample is actually selected is called the sampling frame. Ideally, the frame should list every individual in the population, but in practice this is often difficult. A frame that leaves out part of the population is a common source of undercoverage.
Suppose that a sample of households in a community is selected at random from the telephone directory.
What types of people do you think are likely to live in these households?These people will probably be underrepresented in the sample. (Select all that apply.)1.People with extra time on their hands.2.People without the extra income to keep a phone line will largely comprise the population without a phone line.3.Very busy people or those interested in maintaining privacy will largely comprise those with unlisted numbers.4.People who do not like to take surveys are likely to live in these households.
Answer:
Step-by-step explanation:
answer to Step 1
Households with unlisted numbers or without telephones.
answer to Step 2
2.People without the extra income to keep a phone line will largely comprise the population without a phone line.
3.Very busy people or those interested in maintaining privacy will largely comprise those with unlisted numbers.
A person walks in the following pattern: 2.4 km north, then 1.9 km west, and finally 4.7 km south. (a) How far and (b) at what angle (measured counterclockwise from east) would a bird fly in a straight line from the same starting point to the same final point?
Answer:
a) Figure attached
b) For this case w have that A =(2.4 km)j, B= (-1.9 km) i , C= (-4.7 km)j
And the final position vector can be calculated adding the 3 vectors like this:
[tex] s = A +B+C[/tex]
[tex] s= (-1.9 km)i +(2.4 -4.7 km) j= (-1.9km)i + (-2.3 km)j[/tex]
We can find the magnitude of s like this:
[tex] |s| = \sqrt{(-1.9)^2 +(-2.3)^2}=2.983[/tex]
And then we can find the angle with this formula:
[tex] \theta = \tan^{-1} (\frac{-2.3 km}{-1.9 km})=50.44 [/tex]
The other possibility is [tex] \theta = 50.44+180 =230.44[/tex]
And since they want the angle measured from East the correct angle would be [tex] \theta = 230.44[/tex]
Step-by-step explanation:
Part a
On the figure attached we have the vectors for the pattern described.
Part b
For this case w have that A =(2.4 km)j, B= (-1.9 km) i , C= (-4.7 km)j
And the final position vector can be calculated adding the 3 vectors like this:
[tex] s = A +B+C[/tex]
[tex] s= (-1.9 km)i +(2.4 -4.7 km) j= (-1.9km)i + (-2.3 km)j[/tex]
We can find the magnitude of s like this:
[tex] |s| = \sqrt{(-1.9)^2 +(-2.3)^2}=2.983[/tex]
And then we can find the angle with this formula:
[tex] \theta = \tan^{-1} (\frac{-2.3 km}{-1.9 km})=50.44 [/tex]
The other possibility is [tex] \theta = 50.44+180 =230.44[/tex]
And since they want the angle measured from East the correct angle would be [tex] \theta = 230.44[/tex]
A prize was awarded to 56 women and 642 men.
a. What fraction of the prize winners were women?
b. What fraction were men?
SIMPLIFY EACH plz
At a certain college, 30% of the students major in engineering, 20% play club sports, and 10% both major in engineering and play club sports. A student is selected at random. What is the probability that the student is majoring in engineering?
The probability that a student selected at random majors in engineering is 30% which is 0.3.
The probability that the student both majors in engineering and play club sports is 10% which is 0.1.
For a student who is selected at random to be one who majors in engineering, there are two possible ways.
The student majors in engineering OR the Student both majors in engineering and plays club
The Probabilty that the student majors in Enginnering =
The probability that the student majors in engineering plus the probability that Student both majors in engineering and plays club sport
= 0.3+0.1= 0.4
Find the solution of the system of equations 3x+4y=10 and x−y=1. Give the x value followed by the y value, separated by a comma
Answer:
Solution of the system is (2,1).
Step-by-step explanation:
We are given the following system of equation:
[tex]3x+4y=10\\x - y = 1[/tex]
We would use the elimination method to solve the following system of equation.
Multiplying the second equation by 4 and adding the two equation we gwt:
[tex]3x + 4y = 10\\4\times (x-y = 1)\\\Rightarrow 4x - 4y = 4\\\text{Adding equations}\\3x + 4y + (4x-4y) = 10 + 4\\7x = 14\\\Rightarrow x = 2\\\text{Substituting value of x in second equation}\\2 - y = 1\\\Rightarrow y = 1[/tex]
Solution of the system is (2,1).
Answer: the solution is (2, 1)
Step-by-step explanation:
The given system of simultaneous equations is given as
3x+4y=10 - - - - - - - - - - - - -1
x−y=1 - - - - - - - - - - - - 2
We would eliminate x by multiplying equation 1 by 1 an equation 2 by 3. It becomes
3x + 4y = 10
3x - 3y = 3
Subtracting, it becomes
7y = 7
Dividing the left hand side and the right hand side of the equation by 7, it becomes
7y/7 = 7/7
y = 1
Substituting y = 1 into equation 2, it becomes
x - 1 = 1
Adding 1 to the left hand side and the right hand side of the equation, it becomes
x - 1 + 1= 1 + 1
x = 2
If 400 mL of a 20% w/v solution were diluted to 2 L, what would be the final percentage strength?
Answer:
4%
Step-by-step explanation:
A 20% solution with a total volume of 400 mL has 20% * 400 mL of solute.
20% * 400 mL = 80 mL
When you dilute the solution to 2 L, you introduce additional water, but no additional solute, so now you have the same 80 mL of solute in 2 L of total solution.
The concentration is:
(80 mL)/(2 L) = (80 mL)/(2000 mL) = 0.04
As a percent it is:
0.04 * 100% = 4%
"4" will be the final percentage strength.
According to the question,
20% solution with 400 mL has 20% × 400 mL of solute
then,
→ [tex]20 \ percent\times 400=80[/tex]
hence,
The concentration will be:
→ [tex]\frac{80 \ mL}{2 \ L} = \frac{80 \ mL}{2000 \ mL}[/tex]
[tex]= 0.04[/tex]
or,
[tex]= 0.04\times 100[/tex]
[tex]= 4[/tex] (%)
Thus the answer above is right.
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Without solving the equation 3x2−12x+7=0 find the sum of the squares of its roots.
Answer:
Step-by-step explanation:
3x²-12x+7=0
x = (-b ± √ b²-4ac) / 2a
from these equation: a = 3 , b = 12 , c = 7
Accordingly, b² - 4ac = 12²- 4 (3 x 7) = 144 - 4(21) = 144 - 84 = 60
2a= 2 x 3 = 6
Applying the quadratic formula :
x = (-12 ± √ 60 )/6
√ 60 rounded to 2 decimal digit using calculator 7.75
x = (-12 ± 7.75 )/6
(-12 ± 7.75 ) = (-12 + 7.75 ) = 5.75
(-12 - 7.75 ) = -19.75
x = 5.75/6 = 0.96
-19.75/6 = 3.29
sum of the square of roots = 0.96² + 3.29²
0.9261 + 10.8241 = 11.75
The sum of the squares of its roots of the quadratic equation
3x² - 12x + 7 = 0 is p² + q² = 34/3.
What is a quadratic equaton?A quadratic equation is an algebraic expression in the form of variables and constants.
A quadratic equation has two roots as its degree is two.
We have a quadratic equation 3x² - 12x + 7 = 0.
We know a quadratic equation ax² - bx + c = 0 can be written as,
x² - (b/a)x + (c/a) = 0 ⇒ x² - (p + q)x + pq = 0 where p and q are the roots.
∴ 3x² - 12x + 7 = 0.
x² - (12/3)x + 7/3 = 0.
Hence (p + q) = 4 and pq = 7/3.
Now, (p + q)² = p² + q² + 2pq.
16 = p² + q² + 14/3.
p² + q² = 16 - 14/3.
p² + q² = (48 - 14)/3.
p² + q² = 34/3.
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Sarah inherited a sum of 1,539 dollars with an interest rate of 1% per year what is the accrued value after 5 years, 10 years, and 15 years. Show all work.
Answer:
The accrued value after 5 years is $1,605.95.
The accrued value after 10 years is $1,672.9.
The accrued value after 15 years is $1,739.85.
Step-by-step explanation:
This is a simple interest problem.
The simple interest formula is given by:
[tex]E = P*I*t[/tex]
In which E are the earnings, P is the principal(the initial amount of money), I is the interest rate(yearly, as a decimal) and t is the time.
After t years, the total amount of money is:
[tex]T = E + P[/tex].
In this problem, we have that:
[tex]P = 1539, I = 0.01[/tex]
Accrued value after 5 years
This T when t = 5. So
[tex]E = P*I*t[/tex]
[tex]E = 1339*0.01*5 = 66.95[/tex]
The total is
[tex]T = E + P = 66.95 + 1539 = 1605.95[/tex]
The accrued value after 5 years is $1,605.95.
Accrued value after 10 years
This T when t = 10. So
[tex]E = P*I*t[/tex]
[tex]E = 1339*0.01*10 = 133.9[/tex]
The total is
[tex]T = E + P = 133.9 + 1539 = 1672.9[/tex]
The accrued value after 10 years is $1,672.9.
Accrued value after 15 years
This T when t = 15. So
[tex]E = P*I*t[/tex]
[tex]E = 1339*0.01*15 = 200.85[/tex]
The total is
[tex]T = E + P = 200.85 + 1539 = 1739.85[/tex]
The accrued value after 15 years is $1,739.85.
New drug strategies have centered on cholesterol ______ statins. Select one:
a. lowering b. raising c. destroying d. improving
Answer: lowering
Step-by-step explanation:
Statins are drugs that are administered in order to help reduce cholesterol level in the body. Statins achieves this by making it impossible for the body to have access to substances that helps to produce cholesterol. This drug reduces cholesterol up to 50% and are taken once every day. Examples includes rosuvastatin, pitavastatin, simvastatin.
When 258 college students are randomly selected and surveyed, it is found that 106 own a car. Find a 99% confidence interval for the true proportion of all college students who own a car. 0.351 < p < 0.471 0.332 < p < 0.490 0.339 < p < 0.482 0.360 < p < 0.461
Answer: 0.332 < p < 0.490
Step-by-step explanation:
We know that the confidence interval for population proportion is given by :-
[tex]\hat{p}\pm z^*\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex]
, where n= sample size
[tex]\hat{p}[/tex] = sample proportion
z* = critical z-value.
As per given , we have
n= 258
Sample proportion of college students who own a car = [tex]\hat{p}=\dfrac{106}{258}\approx0.411[/tex]
Critical z-value for 99% confidence interval is 2.576. (By z-table)
Therefore , the 99% confidence interval for the true proportion(p) of all college students who own a car will be :[tex]0.411\pm (2.576)\sqrt{\dfrac{0.411(1-0.411)}{258}}\\\\=0.411\pm (2.576)\sqrt{0.00093829}\\\\= 0.411\pm (2.576)(0.0306315197142)\\\\=0.411\pm 0.0789=(0.411-0.0789,\ 0.411+0.0789)\\\\=(0.3321,\ 0.4899)\approx(0.332,\ 0.490)[/tex]
Hence, a 99% confidence interval for the true proportion of all college students who own a car : 0.332 < p < 0.490
which is the equation of a libe that has a slope of 1/2 and passes through (2,-3)
Answer:
Step-by-step explanation:
The equation of a straight line can be represented in the slope-intercept form, y = mx + c
Where c = intercept
Slope, m =change in value of y on the vertical axis / change in value of x on the horizontal axis
change in the value of y = y2 - y1
Change in value of x = x2 -x1
The slope is given as 1/2 and the line passes through (2, - 3)
To determine the intercept, we would substitute x = 2, y = - 3 and m= 1/2 into y = mx + c
y = mx + c. It becomes
- 3 = 1/2 × 2 + c = 1 + c
c = - 3 - 1 = - 4
The equation becomes
y = x/2 - 4
Let X and Y be independent random variables representing the lifetime (in 100 hours) of Type A and Type B light bulbs, respectively.
Both variables have exponential distributions, and the mean of X is 2 and the mean of Y is 3.
Find the joint pdf f(x|y) of X and Y.
Find the conditional pdf f_2(y|x) of Y given X = x.
Find the probability that a Type A bulb lasts at least 300 hours and a Type B bulb lasts at least 400 hours.
Given that a Type B bulb fails at 300 hours, find the probability that a Type A bulb lasts longer than 300 hours.
What is the expected total lifetime of two Type A bulbs and one Type B bulb?
What is the variance of the total lifetime of two Type A bulbs and one Type B bulb?
Answer:
a. is below in the explanation
b. 1/3e^(-y/3)
c.0.05882
d. 0.2231
e. 7 f.25
Step-by-step explanation:
Let and be independent random variables representing the lifetime (in 100 hours) of Type A and Type B light bulbs, respectively. Both variables have exponential distributions, and the mean of X is 2 and the mean of Y is 3.
a can be solved as follows [tex]\frac{1}{\alpha\beta } e^{\frac{-x-y}{\alpha\beta } }[/tex]
[tex]\frac{1}{6} e^{\frac{-\alpha }{2} -\beta/3 }[/tex]
1/6[tex]e^{\frac{-(3x+2y}{6 }[/tex]
b.
f(y/x)=f(x).f(y)/{f(y)}=f(y)
1/3e^(-y/3)
c. Find the probability that a Type A bulb lasts at least 300 hours and a Type B bulb lasts at least 400 hours.
[tex]\int\limits^\alpha _4 {1/6e^{-(3x+2y)/6} } \, dy[/tex]
1/2e^-(3x+8)/6
[tex]\int\limits^\alpha _3 {e^{-(3x+800)} } \, dx \\[/tex]
0.05882
Given that a type B fails at 300 hours . we find the probability that type A bulb lasts longer than 300hr
f(x>y|y=3)=f(x>3)
[tex]\int\limits^a_3 {1/2e^{-x/2} } \, dx[/tex]
0.2231
e.e) What is the expected total lifetime of two Type A bulbs and one Type B bulb?
E(2A+B)
(2E(A)+E(y)
2*2+3
=7
f. variance of the total lifetime of two types A bulbs and one type B bulb
V(2x+y)=
4*4+9
=25
The joint pdf of X and Y is (1/6)e^(-x/2-y/3). The conditional pdf of Y given X = x is (1/2)e^(-y/3). The probability that a Type A bulb lasts at least 300 hours and a Type B bulb lasts at least 400 hours is 9/250. The probability that a Type A bulb lasts longer than 300 hours, given that a Type B bulb fails at 300 hours, is approximately 0.9999. The expected total lifetime of two Type A bulbs and one Type B bulb is 7 100 hours. The variance of the total lifetime of two Type A bulbs and one Type B bulb is 10 100 hours.
Explanation:To find the joint pdf f(x|y) of X and Y, we need to find the product of the individual pdfs of X and Y since they are independent. Since X and Y follow exponential distributions with means of 2 and 3 respectively, their pdfs can be written as f(x) = (1/2)e^(-x/2) and f(y) = (1/3)e^(-y/3). Therefore, the joint pdf f(x|y) is given by:
f(x|y) = f(x)*f(y) = (1/2)e^(-x/2) * (1/3)e^(-y/3) = (1/6)e^(-x/2-y/3).
To find the conditional pdf f_2(y|x) of Y given X = x, we use Bayes' theorem:
f_2(y|x) = (f(x|y)*f(y))/f(x) = [(1/6)e^(-x/2-y/3) * (1/3)e^(-y/3)] / [(1/2)e^(-x/2)] = (1/2)e^(-y/3).
To find the probability that a Type A bulb lasts at least 300 hours and a Type B bulb lasts at least 400 hours, we integrate the joint pdf over the given range:
P(X >= 300, Y >= 400) = ∫∫(1/6)e^(-x/2-y/3)dxdy = ∫(3/2)e^(-x/2)dx * ∫e^(-y/3)dy. Solving the integrals, we get P(X >= 300, Y >= 400) = 9/250.
Given that a Type B bulb fails at 300 hours, we need to find the probability that a Type A bulb lasts longer than 300 hours. This is equivalent to finding P(X > 300 | Y = 300). Using the conditional pdf f_2(y|x) = (1/2)e^(-y/3), we integrate from 300 to infinity:
P(X > 300 | Y = 300) = ∫(1/2)e^(-y/3)dy = 1 - ∫(1/2)e^(-y/3)dy = 1 - (1/2)e^(-y/3) = 1 - (1/2)e^(-300/3) = 1 - e^(-100) ≈ 0.9999.
The expected total lifetime of two Type A bulbs and one Type B bulb is obtained by summing the means of X and Y twice and the mean of Y once:
E[Total Lifetime] = 2*E[X] + E[Y] = 2*2 + 3 = 7.
The variance of the total lifetime of two Type A bulbs and one Type B bulb is obtained by summing the variances of X and Y twice and the variance of Y once:
Var[Total Lifetime] = 2*Var[X] + Var[Y] = 2*(2^2) + 3^2 = 10.
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Let S = {2,4,6} and T = {1,3,5}. Use the set-roster notation to write each of the following sets, and indicate the number of elements that are in each set?
a. S X T
b. T X S
c. S X S
d. T X T
Answer:
a. [tex]S\times T=\{(2,1),(2,3),(2,5),(4,1),(4,3),(4,5),(6,1),(6,3),(6,5)\}[/tex] . Set S × T has 9 elements.
b. [tex]T\times S=\{(1,2),(1,4),(1,6),(3,2),(3,4),(3,6),(5,2),(5,4),(5,6)\}[/tex]. Set T × S has 9 elements.
c. [tex]S\times S=\{(2,2),(2,4),(2,6),(4,2),(4,4),(4,6),(6,2),(6,4),(6,6)\}[/tex] . Set S × S has 9 elements.
d. [tex]T\times T=\{(1,1),(1,3),(1,5),(3,1),(3,3),(3,5),(5,1),(5,3),(5,5)\}[/tex] . Set T × T has 9 elements.
Step-by-step explanation:
The given sets are S={2,4,6} and T={1,3,5}.
We need to find the set-roster notation to write each of the following sets, and the number of elements that are in each set.
a.
[tex]S\times T=\{(s,t)|s\in S and t\in T\}[/tex]
[tex]S\times T=\{(2,1),(2,3),(2,5),(4,1),(4,3),(4,5),(6,1),(6,3),(6,5)\}[/tex]
Set S × T has 9 elements.
b.
[tex]T\times S=\{(t,s)|s\in S and t\in T\}[/tex]
[tex]T\times S=\{(1,2),(1,4),(1,6),(3,2),(3,4),(3,6),(5,2),(5,4),(5,6)\}[/tex]
Set T × S has 9 elements.
c.
[tex]S\times S=\{(s,s)|s\in S \}[/tex]
[tex]S\times S=\{(2,2),(2,4),(2,6),(4,2),(4,4),(4,6),(6,2),(6,4),(6,6)\}[/tex]
Set S × S has 9 elements.
d.
[tex]T\times T=\{(t,t)|t\in T\}[/tex]
[tex]T\times T=\{(1,1),(1,3),(1,5),(3,1),(3,3),(3,5),(5,1),(5,3),(5,5)\}[/tex]
Set T × T has 9 elements.
Using the set roster notation, the number of elements in each of the sets would be the product of the number of elements in each individual set, which is 9.
S = {2, 4, 6}T = {1, 3, 5}A.) S × T :
{2, 4, 6} × {1, 3, 5}
{(2, 1), (2, 3), (2, 5), (4, 1), (4, 3), (4, 5), (6, 1), (6, 3), (6, 5)}
Number is of elements = 9
B.) T × S :
{1, 3, 5} × {2, 4, 6}
{(1,2), (1,4), (1,6), (3,2), (3,4), (3, 6), (5, 2), (5,4), (5,6)}
Number is of elements = 9
C.) S × S :
{1, 3, 5} × {1, 3, 5}
{(1, 1), (1, 3), (1, 5), (3, 1), (3, 3), (3, 3), (5, 1), (5, 3), (5, 5)}
Number is of elements = 9
D.) T × T :
{2, 4,6} × {2, 4, 6}
{(2,2), (2, 4), (2, 6), (4,2), (4, 4), (4, 6), (6,2), (6, 4), (6,6)}
Number is of elements = 9
Hence, the distribution of values of each set.
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the absolute value of a complex number z=a+bi, denoted |z|, is given by the formula
|z|=square root a^2+b^2.
find six distinct complex numbers (which are neither purely imaginary nor purley real) such that each has an absolute value of 7.
Step-by-step explanation:
You need to find six combinations of a and b such that a² + b² = 7.
a = 1, b = √6
a = √2, b = √5
a = √3, b = 2
a = 2, b = √3
a = √5, b = √2
a = √6, b = 1
Find the Surface area
Answer:
27m²
Step-by-step explanation:
surface area is area of all sides
these is two triangles and a rectangle
area of triangle = (base x height) /2 = (3 x 4)/2= 12/2 =6
area of rectangle = length x breadth = 5x3 = 15
surface area = area of triangle + area of triangle + area of rectangle
6 + 6 + 15 = 27m²
Charlie is a car salesman. He earns $800 plus a 9% commission on all of his sales for the week. If Charlie received a paycheck for 1,916 this week, how much were his total sales?
Answer:
$12,400
Step-by-step explanation:
Given:
Week earning + 9% commission = $1,916
Weekly Earning = $800
Commission for the Week = $1,916 - $800 = $1,116
Let total sales for the week = x
Therefore, x = ($1,116 * 100)/9
x = $12,400
Answer:
$12400
Step-by-step explanation:
Wages = Basic pay + Commission
1916 = 800 + 9%A
Let A be the total sales
9%A = 1916 - 800
9%A = 1116
A = $12400
help pls asap in Ixl for geometry
Answer:
Step-by-step explanation:
Triangle QRS is a right angle triangle.
From the given right angle triangle
RS represents the hypotenuse of the right angle triangle.
With 30 degrees as the reference angle,
QR represents the adjacent side of the right angle triangle.
QS represents the opposite side of the right angle triangle.
To determine QR, we would apply trigonometric ratio
Cos θ = adjacent side/hypotenuse side. Therefore,
Cos 30 = QR/14
√3/2 = QR/14
QR = 14 × √3/2
QR = 14√3/2 = 7√3/2
Which of the following equations is also true?
Answer:
cos 38 = 17/c
Step-by-step explanation:
cos 38 = 17/c is the correct statement
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The cost, c, of a ham sandwich at a deli varies directly with the number of sandwiches, n. If c = $54 when n is 9, what is the
cost of the sandwiches when n is 3?
$18
$21
$27
$48
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Answer:
$18
Step-by-step explanation:
The cost, c, of a ham sandwich at a deli varies directly with the number of sandwiches, n. c = $54 when n is 9
This means that to buy 9 sandwiches, it costs $54. So one sandwich costs 54/9 = $6.
What is the cost of the sandwiches when n is 3?
To buy three sandwiches, it costs $6*3 = $18.
So the correct answer is:
$18
Answer:A
Step-by-step explanation:
$18
uppose that a certain fortunate person has a net worth of $71.0 billion ($7.10×1010). If her stock has a good year and gains $3.20 billion (3.20×109) in value, what is her new net worth
The new net worth after a gain in stock value is $74.2 billion, or in the scientific notation it is $7.42×10¹0.
Explanation:In this problem, we have a person with a starting net worth of $71.0 billion, or $7.10×1010 in scientific notation. This person's stock has increased by $3.20 billion, or 3.20×109 in scientific notation. To find the new net worth, these two amounts should be added together.
Step 1: Write the starting net worth and stock increase in standard form: $71.0×10⁹ + $3.2×10⁹. Step 2: Add these two amounts together to find the new net worth. $74.2×10⁹ This is the new net worth in standard form.
To express this amount in scientific notation, it becomes $7.42×10¹0, which is your final answer.
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Find the matrix P that multiplies (x; y; z) to give (y; z; x). Find the matrix Q that multiplies (y; z; x) to bring back (x; y; z).
Answer:
[tex]P = \left[\begin{array}{ccc}0&1&0\\0&0&1\\1&0&0\end{array}\right][/tex]
[tex]Q = \left[\begin{array}{ccc}0&0&1\\1&0&0\\0&1&0\end{array}\right][/tex]
Step-by-step explanation:
[tex]P*\left[\begin{array}{c}x\\y\\z\end{array}\right]=\left[\begin{array}{ccc}0&1&0\\0&0&1\\1&0&0\end{array}\right] \left[\begin{array}{c}x\\y\\z\end{array}\right] = \left[\begin{array}{c}y\\z\\x\end{array}\right][/tex]
[tex]Q*\left[\begin{array}{c}y\\z\\x\end{array}\right]=\left[\begin{array}{ccc}0&0&1\\1&0&0\\0&1&0\end{array}\right] \left[\begin{array}{c}y\\z\\x\end{array}\right] = \left[\begin{array}{c}x\\y\\z\end{array}\right][/tex]
A quantity and its 2/3 are added together and from thesum 1/3 of the sum is subtracted, and 10 remains.What is the quantity?
Answer:
[tex]14\frac{1}{3}[/tex]
Step-by-step explanation:
Let's write this out as an equation. Let the unknown quantity be x:
[tex]x+\frac{2}{3} - \frac{1}{3} (x+\frac{2}{3}) = 10\\\\3x+2-x -\frac{2}{3} = 30\\\\9x+6-3x-2=90\\\\6x= 86\\\\x = \frac{86}{6} =\frac{43}{3} =14\frac{1}{3} \\[/tex]
Answer: the quantity is 9
Step-by-step explanation:
Let x represent the quantity.
A quantity and its 2/3 are added together. The 2/3 of the number is 2/3 × x = 2x/3
The sum of the quantity and its 2/3 would be
x + 2x/3 = (3x + 2x)/3 = 5x/3
From the sum, 1/3 of the sum is subtracted. 1/3 of the sum would be
1/3 × 5x/3 = 5x/9
Subtracting 1/3 of the sum from the sum, it becomes
5x/3 - 5x/9 = (15x - 5x)/9 = 10x/9
If the remainder is 10, it means that
10x/9 = 10
Crossmultiplying
10x = 9 × 10 = 90
x = 90/10
x = 9
Solve the right triangle
SR =sin (angle) = opposite leg/ hypotenuse
sin(52) = SR/ 7.6
SR = 7.6 x sin(52)
SR = 5.988 ( round answer as needed.)
QR = cos(angle) = adjacent leg / hypotenuse
Cos(52) = QR/ 7.6
QR = 7.6 x cos(52)
QR = 4.679. ( round answer as needed).
Angle S = 180 - 90 -52 = 38 degrees
USA Today reported that about 20% of all people in the United States are illiterate. Suppose you take eleven people at random off a city street. (a) Make a histogram showing the probability distribution of the number of illiterate people out of the eleven people in the sample.
Final answer:
To create a histogram showing the probability distribution of the number of illiterate people out of eleven people in a sample, you can use the binomial probability formula. Calculate the probability for each possible number of illiterate people and represent it in a histogram.
Explanation:
This question is asking for a histogram showing the probability distribution of the number of illiterate people out of eleven people in a sample, based on the information that about 20% of all people in the United States are illiterate.
To create the histogram, we can use the binomial probability formula. The formula is P(X = k) = C(n, k) * p^k * (1 - p)^(n - k), where C(n, k) represents combinations, n is the number of trials, k is the number of successes, and p is the probability of success.
We can calculate the probability for each possible number of illiterate people out of the eleven, ranging from 0 to 11, and represent it in a histogram.
math answers- the length of segment AB is 3.6. the length of segment AC is 11.8. what is the length of segment BC? write and solve a decimal addition equation and a decimal subtraction equation.
Answer:
answer is AC^2
Step-by-step explanation:
Determine if the statement is true or false. A system in echelon form can have more variables than equations. True False Justify your answer.
Answer:
Yes
Step-by-step explanation:
A system in echelon form can have more variables than equations. ... Every linear system with free variables has infinitely many solutions. True, a free variable can take any value, and so there are infinitely many solutions. Any linear system with more variables than equations cannot have a unique solution.
What is the difference between relative frequency and cumulative frequency?
A. Relative frequency of a class is the percentage of the data that falls in that class, while cumulative frequency of a class is the sum of the frequencies of that class and all previous classes.
B. Relative frequency of a class is the proportion of the data in that class, while cumulative frequency of a dass is the number of observations in that class.
C. Relative frequency of a class is the number of observations in that dass, while cumulative frequency of a dass is the sum of all the frequencies.
D. There is no difference between the two.
Answer:
A
Step-by-step explanation:
The relative frequency is calculated by dividing the frequency of that class to the sum of frequencies. It can be represented as
[tex]Relative frequency=\frac{f}{sum(f)}[/tex]
Hence, the relative frequency of class is the percentage or proportion of data lies in that class.
The cumulative frequency of a class is computed by adding the frequency of the respective class to the frequencies of all previous classes. The cumulative frequency of first class will always be equal to the frequency of first class.
Hence, the cumulative frequency for a class is the sum of frequency for that class and the frequencies of all previous classes.
Relative frequency is the proportion of total data that falls into a class, while cumulative frequency is the sum of frequencies of that class and all previous classes. Therefore, relative frequency illustrates the percentage of data in a certain class, while cumulative frequency shows the accumulation of data up to that point.
Explanation:The terms relative frequency and cumulative frequency both pertain to statistics; however, they represent different concepts. Relative frequency of a class is the percentage or proportion of the whole set of data that falls in that class. It's calculated by dividing the frequency of that class by the total number of data points.
On the other hand, cumulative frequency of a class is the sum of the frequencies of that class and all previous classes in a dataset. Essentially, it accumulates counts as you proceed through the dataset.
For instance, if you have five classes with frequencies of 1, 2, 3, 4 and 5, the relative frequencies would be 0.0667, 0.1333, 0.2, 0.2667, and 0.3333 respectively (assuming we divide each frequency by the total data points, which is 15 in this case). In contrast, the cumulative frequencies would be 1, 3, 6, 10 and 15, indicating the total count up to each class.
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The table below shows selected points from a function.
The rate of change for the interval shown in the table is
A. constant
B. not constant
part b
A. linear
B. not linear
The rate of change is A. constant, B. linear
Step-by-step explanation:
Rate of change is the ratio of change in value of y with corresponding value of x
Rate of change=Δy/Δx
Rate of change=5-2/4-1 =3/3=1
The rate of change is constant in the given interval and linear with a positive slope.
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What is the simple interest on a loan of $16 500 for 2 years at 8 per year ?
Answer:
$2 640
Step-by-step explanation:
Principal = $16 500; Time = 2 years Rate = 8%
Simple Interest = P × R × T/100
= $16 500 × 8 × 2/100
=$264 000/100
Simple Interest = $2 640